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Karnataka Board PUCPUC Science Class 11

A Wheel of Moment of Inertia 0⋅500 Kg-m2 and Radius 20⋅0 Cm is Rotating About Its Axis at an Angular Speed of 20⋅0 Rad/S.

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Question

A wheel of moment of inertia 0⋅500 kg-m2 and radius 20⋅0 cm is rotating about its axis at an angular speed of 20⋅0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

Sum
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Solution

Given

Initial moment of inertia of the system,

I1 = 0.500 kg-m2;

r = 0.2 m;

ω = 20 rad/s

Mass of the stationary particle, m = 0.2 kg

Final moment of inertia of the system,

I2 = I1 + mr2

It is given

External torque = 0

Angular momentum is conserved; therefore, we have

\[I_1  \omega_1  =  I_2  \omega_2\]

\[\Rightarrow 0 . 5 \times 20 = \left( 0 . 5 + 0 . 2 \times \left( 0 . 2 \right)^2 \right)   \omega_2 \] 

\[ \Rightarrow  \omega_2  = \frac{10}{0 . 508} \approx 19 . 7\text{ rad/s}\]

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Chapter 10: Rotational Mechanics - Exercise [Page 198]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 10 Rotational Mechanics
Exercise | Q 51 | Page 198

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